BZOJ 1114 Number theory（莫比乌斯反演+预处理）

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <algorithm>
 4 using namespace std;
 5 typedef long long LL;
 6 const int MAXN = 250000;
 7 int num[MAXN];        // num[i]表示 满足(i|ak)的个数 
 8 int tmp[MAXN];        // 标记哪些数有
 9 bool check[MAXN];
10 int primes[MAXN];
11 int mu[MAXN];
12 
13 void moblus()
14 {
15     memset(check, false, sizeof(check));
16     mu[1] = 1;
17     int cnt = 0;
18     for (int i = 2; i < MAXN; ++i) {
19         if (!check[i]) {
20             primes[cnt++] = i;
21             mu[i] = -1;
22         }
23         for (int j = 0; j < cnt; ++j) {
24             if (i * primes[j] > MAXN) break;
25             check[i * primes[j]] = true;
26             if (i % primes[j] == 0) {
27                 mu[i * primes[j]] = 0;
28                 break;
29             } else {
30                 mu[i * primes[j]] = -mu[i];
31             }
32         }
33     }
34 }
35 
36 int main()
37 {
38     moblus();
39     int n;
40     while (scanf("%d", &n) != EOF) {
41         memset(num, 0, sizeof(num));
42         memset(tmp, 0, sizeof(tmp));
43         int tmax = 0;
44         for (int i = 0; i < n; ++i) {
45             int x;
46             scanf("%d", &x);
47             ++tmp[x];
48             tmax = max(tmax, x);
49         }
50         for (int i = 1; i <= tmax; ++i) {
51             for (int j = i; j <= tmax; j += i) {
52                 num[i] += tmp[j];
53             }
54         }
55         LL ans = 0;
56         for (int i = 1; i <= tmax; ++i) {
57             ans += (LL)mu[i] * num[i] * (num[i] - 1) / 2;
58         }
59         printf("%lld\n", ans);
60     }
61     return 0;
62 }